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E = | F q | = k | qQ qr 2 | = k | Q | r 2 . size 12{E= { {F} over {q} } =k { { ital "qQ"} over { ital "qr" rSup { size 8{2} } } } =k { {Q} over {r rSup { size 8{2} } } } } {}

Since the test charge cancels, we see that

E = k | Q | r 2 . size 12{E=k { {Q} over {r rSup { size 8{2} } } } } {}

The electric field is thus seen to depend only on the charge Q size 12{Q} {} and the distance r size 12{r} {} ; it is completely independent of the test charge q size 12{q} {} .

Calculating the electric field of a point charge

Calculate the strength and direction of the electric field E size 12{E} {} due to a point charge of 2.00 nC (nano-Coulombs) at a distance of 5.00 mm from the charge.

Strategy

We can find the electric field created by a point charge by using the equation E = kQ / r 2 size 12{E= { ital "kQ"} slash {r rSup { size 8{2} } } } {} .

Solution

Here Q = 2 . 00 × 10 9 size 12{Q=2 "." "00" times "10" rSup { size 8{ - 9} } } {} C and r = 5 . 00 × 10 3 size 12{r=5 "." "00" times "10" rSup { size 8{ - 3} } } {} m. Entering those values into the above equation gives

E = k Q r 2 = ( 8.99 × 10 9 N m 2 /C 2 ) × ( 2.00 × 10 9 C ) ( 5.00 × 10 3 m ) 2 = 7.19 × 10 5 N/C. alignl { stack { size 12{E=k { {Q} over {r rSup { size 8{2} } } } } {} #= \( 9 "." "00" times "10" rSup { size 8{9} } N cdot m rSup { size 8{2} } "/C" rSup { size 8{2} } \) times { { \( 2 "." "00" times "10" rSup { size 8{ - 9} } C \) } over { \( 5 "." "00" times "10" rSup { size 8{ - 3} } m \) rSup { size 8{2} } } } {} # =7 "." "20" times "10" rSup { size 8{5} } "N/C" {}} } {}

Discussion

This electric field strength is the same at any point 5.00 mm away from the charge Q size 12{Q} {} that creates the field. It is positive, meaning that it has a direction pointing away from the charge Q size 12{Q} {} .

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Calculating the force exerted on a point charge by an electric field

What force does the electric field found in the previous example exert on a point charge of –0.250 μ C ?

Strategy

Since we know the electric field strength and the charge in the field, the force on that charge can be calculated using the definition of electric field E = F / q size 12{E= {F} slash {q} } {} rearranged to F = q E size 12{F= ital "qE"} {} .

Solution

The magnitude of the force on a charge q = 0 . 250 μC size 12{q= - 0 "." "250""μC"} {} exerted by a field of strength E = 7 . 20 × 10 5 size 12{E=7 "." "20" times "10" rSup { size 8{5} } } {} N/C is thus,

F = qE = ( 0.250 × 10 –6 C ) ( 7.20 × 10 5 N/C ) = 0.180 N. alignl { stack { size 12{F= ital "qE"} {} #size 12{ {}= \( "-0" "." "250" times "10" rSup { size 8{"-6"} } `C \) \( 7 "." "20" times "10" rSup { size 8{5} } `"N/C" \) } {} # ="-0" "." "180"`N {}} } {}

Because q is negative, the force is directed opposite to the direction of the field.

Discussion

The force is attractive, as expected for unlike charges. (The field was created by a positive charge and here acts on a negative charge.) The charges in this example are typical of common static electricity, and the modest attractive force obtained is similar to forces experienced in static cling and similar situations.

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Phet explorations: electric field of dreams

Play ball! Add charges to the Field of Dreams and see how they react to the electric field. Turn on a background electric field and adjust the direction and magnitude.

Electric Field of Dreams

Section summary

  • The electrostatic force field surrounding a charged object extends out into space in all directions.
  • The electrostatic force exerted by a point charge on a test charge at a distance r size 12{r} {} depends on the charge of both charges, as well as the distance between the two.
  • The electric field E size 12{E} {} is defined to be
    E = F q , size 12{E= { {F} over {q,} } } {}

    where F size 12{F} {} is the Coulomb or electrostatic force exerted on a small positive test charge q size 12{q} {} . E size 12{E} {} has units of N/C.

  • The magnitude of the electric field E size 12{E} {} created by a point charge Q size 12{Q} {} is
    E = k | Q | r 2 . size 12{E=k { {Q} over {r rSup { size 8{2} } } } } {}

    where r size 12{r} {} is the distance from Q size 12{Q} {} . The electric field E size 12{E} {} is a vector and fields due to multiple charges add like vectors.

Conceptual questions

Why must the test charge q size 12{q} {} in the definition of the electric field be vanishingly small?

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Are the direction and magnitude of the Coulomb force unique at a given point in space? What about the electric field?

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Problem exercises

What is the magnitude and direction of an electric field that exerts a 2 . 00 × 10 - 5 N size 12{2 "." "00" times "10" rSup { size 8{5} } N} {} upward force on a –1.75 μ C charge?

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What is the magnitude and direction of the force exerted on a 3.50 μ C charge by a 250 N/C electric field that points due east?

8 . 75 × 10 4 size 12{8 "." "75" times "10" rSup { size 8{ - 4} } } {} N

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Calculate the magnitude of the electric field 2.00 m from a point charge of 5.00 mC (such as found on the terminal of a Van de Graaff).

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(a) What magnitude point charge creates a 10,000 N/C electric field at a distance of 0.250 m? (b) How large is the field at 10.0 m?

(a) 6 . 94 × 10 8 C size 12{ {underline {6 "." "94" times "10" rSup { size 8{ - 8} } " C"}} } {}

(b) 6 . 25 N/C size 12{ {underline {6 "." "25"" N/C"}} } {}

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Calculate the initial (from rest) acceleration of a proton in a 5 . 00 × 10 6 N/C size 12{5 "." "00" times "10" rSup { size 8{6} } "N/C"} {} electric field (such as created by a research Van de Graaff). Explicitly show how you follow the steps in the Problem-Solving Strategy for electrostatics.

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(a) Find the direction and magnitude of an electric field that exerts a 4 . 80 × 10 17 N size 12{4 "." "80" times "10" rSup { size 8{ - "17"} } N} {} westward force on an electron. (b) What magnitude and direction force does this field exert on a proton?

(a) 300 N/C ( east ) size 12{ {underline {"300"" N/C " \( "eas"}} {underline {t \) }} } {}

(b) 4 . 80 × 10 17 N ( east ) size 12{ {underline {4 "." "80" times "10" rSup { size 8{ - "17"} } " N " \( "east" \) }} } {}

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Questions & Answers

calculate the tension of the cable when a buoy with 0.5m and mass of 20kg
Iga Reply
what is displacement
Nyamza Reply
what is the meaning of physics
Alausa Reply
to study objects in motion and how they interact or take part in the natural phenomenon of the universe.
Phill
an object that has a small mass and an object has a large mase have the same momentum which has high kinetic energy
Faith Reply
The with smaller mass
Gift
how
Faith
Since you said they have the same momentum.. So meaning that there is more like an inverse proportionality in the quantities used to find the momentum. We are told that the the is a larger mass and a smaller mass., so we can conclude that the smaller mass had higher velocity as compared to other one
Gift
Mathamaticaly correct
megavado
Mathmaticaly correct :)
megavado
I have proven it by using my own values
Gift
Larger mass=4g Smaller mass=2g Momentum of both=8 Meaning V for L =2 and V for S=4 Now find there kinetic energies using the data presented
Gift
grateful soul...thanks alot
Faith
Welcome
Gift
2 stones are thrown vertically upward from the ground, one with 3 times the initial speed of the other. If the faster stone takes 10 s to return to the ground, how long will it take the slower stone to return? If the slower stone reaches a maximum height of H, how high will the faster stone go
Julliene Reply
30s
Gift
how can i calculate it's height
Julliene
is speed the same as velocity
Faith Reply
no
Nebil
in a question i ought to find the momentum but was given just mass and speed
Faith
just multiply mass and speed then you have the magnitude of momentem
Nebil
Yes
Gift
Consider speed to be velocity
Gift
it worked our . . thanks
Faith
Distinguish between semi conductor and extrinsic conductors
Okame Reply
Suppose that a grandfather clock is running slowly; that is, the time it takes to complete each cycle is longer than it should be. Should you (@) shorten or (b) lengthen the pendulam to make the clock keep attain the preferred time?
Aj Reply
I think you shorten am not sure
Uche
shorten it, since that is practice able using the simple pendulum as experiment
Silvia
it'll always give the results needed no need to adjust the length, it is always measured by the starting time and ending time by the clock
Paul
it's not in relation to other clocks
Paul
wat is d formular for newton's third principle
Silvia
okay
Silvia
shorten the pendulum string because the difference in length affects the time of oscillation.if short , the time taken will be adjusted.but if long ,the time taken will be twice the previous cycle.
FADILAT
discuss under damped
Prince Reply
resistance of thermometer in relation to temperature
Ifeanyi Reply
how
Bernard
that resistance is not measured yet, it may be probably in the next generation of scientists
Paul
Is fundamental quantities under physical quantities?
Igwe Reply
please I didn't not understand the concept of the physical therapy
John Reply
physiotherapy - it's a practice of exercising for healthy living.
Paul
what chapter is this?
Anderson
this is not in this book, it's from other experiences.
Paul
am new in the group
Daniel
please I have probably with calculate please can you please and help me out
John Reply
Sure
Gift
What is Boyce law
Sly Reply
Boyles law states that the volume of a fixed amount of gas is inversely proportional to pressure acting on that given gas if the temperature remains constant which is: V<k/p or V=k(1/p)
FADILAT
how to convert meter per second to kilometers per hour
grace Reply
Divide with 3.6
Mateo
multiply by (km/1000m) x (3600 s/h) -> 3.6
Muhammad
Practice Key Terms 3

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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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